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bauera2

Shortening Mini Light Strand

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I am currently working on a Halloween display and wanted to take an old Mini light strand and make a 5 bulb strands for Eyes and mouth on tombstone. I just cut and spliced and it blows all five bulbs when I plug it in. Can anyone help? Do I need to add a resistor, Capacitor, ect....?

Thanks in advance for any help.
Andrew

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If you take the five bulbs without any modification, the bulbs will burn abnormally bright.

I am not a solderer to be able to tell you what size resister you would need to be able to compensate for the loss of bulbs.

When I ran into this problem last year, I ended up wadding all the excess bulbs into a ball and covered the ball in electrical tape.

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Most mini-light strings are 50 lights or multiples of this. The 50 bulbs are in series, so each bulb gets the line voltage (about 120 volts) divided by the number of bulbs. This gives each bulb about 2.4 volts. That's why you see packages of replacement bulbs labeled "2.5 v".

When you shorten the string and have only 5 bulbs, then the 120-volt line voltage will get divided by 5, giving each bulb 24 volts. Obviously, if you put 24 volts across a 2.5 volt bulb, it will blow!

There are 20-light strings made that take 6-volt bulbs. You could use these, but it would mean that you would have 15 unused bulbs.

Another solution is to use a 12-volt transformer and wire 3 sets of 2 6-volt bulbs. Unfortunately, you wouldn't be able to dim the lights.

Another idea is to use neon pilot lights, or super-size your display and use C-7 bulbs.

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I broke down strings last year and put them on my truck. Steven is right about the bulbs being 2.5 volts , 5 of them to a string work perfect on a 12volt dc system!

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Thanks for all the suggestions. After more thought I will go with C7 Bulbs for eyes only. Found some single scoket no switch ones at Menards.

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Some one can check my math and my figures but this is what I came up with.

2.5voltsand .002A per bulb gives a total of 12.5 volts and .01A so in order to limit your current to .01A and to drop the excess 107.5V Ohms law states:

R=E/I
so:
R=107.5/.01=10750

So a 10Kohm resistor should be fine, you can also find 12Kohm resistors

for the power requirements P=I*E
SO:
P=.01*120=1.2Watts

I would pick a 2Watt resistor to be safe

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I cut the number of bulbs down on a string and then ran only 25% power on that channel. worked great.

Mark

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beaker020 wrote:

2.5voltsand .002A per bulb gives a total of 12.5 volts and .01A so in order to limit your current to .01A and to drop the excess 107.5V Ohms law states:


I think your current figure is off. The 50-light set I have says it takes about 37 watts. This gives I = P / E = 37W / 120V =~ 300mA or .3A. Since it's a single loop, this same current flows through every bulb. If you reduce the current through the bulbs, they will not be very bright.

Therefore, if you really wanted to drop the excess 107V with a resistor, it would be 107 / .3 = 33 Ohms. The power dissipated by this resistor would be 107V * .3A = 32 watts!!!

I don't think you want to use a resistor.

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